A recent hand on Stars big game had one player with ~85% equity on the flop (A9 vs KK on a 99T). They ran it 4 times and A9 only won once. What are the odds of this? I used the binomial distribution to calc for A9:

Win 0 0.05%
Win 1 1.15%
Win 2 9.75%
Win 3 36.85%
Win 4 52.20%
===
Total 100.00%

Is this the right way to look at it?

considering the fact that every time KK sucks out, the amount of outs for it to suck out next time are almost exponentially decreased---using a straightforward method that you have isn't accurate.
--------------------------
could you post the suits of the cards?

You method is a good estimate if you want to know the frequency that that A9 will win X times of the 4 runs over time. No matter how the cards fell, the combined equity for the four runs is whatever the starting equity was. However, I think the A9 would have had more like 89% equity, not 85%. At 89%, he will win only once about 0.47% of the time. His average number of fractional wins over time would be .89*4 = 3.56 times.

That said, with a single trial and a single random permutation of the deck, all five outcomes (0,1,2,3, or 4 fractional wins) are equally likely to occur. The probability only has meaning when we repeat the trial over time (like all probabilities).

The reason I said your method is a good estimate, is that the exact permutations of how the four runs can play out is discrete, and it I think a simulation would be the easiest way to get a more accurate number. I'm pretty sure it won't follow an exact binomial distribution. There are 990 ways the turn and river can fall the first time, 903 the second time, 820 ways the third time, and 741 ways the last time. So with some effort we could just count them.

The suits only matter to get the exact starting equity, but the problem is the same whether that is 85% or 89% or 88.5%. The combined equity of the four runs will be that number no matter how the cards fell.

I realized part of my post above is wrong. This is not correct:
Also, as I mentioned the distribution isn't going to be exactly binomial because the equity in each run changes, although it may be a good estimate. We know the combined equity is 89%, but it isn't the same for all four runs. On the first run, we expect to win ~89% of the 990 possible boards (turn and river) or about 881 of those, and that is for 25% of the pot. So our first run equity is always 22.25% of the pot. But of the 903 possible second runs, it won't be .25*89%, or for the 820 third runs, or for the 741 fourth runs. We know that the combined equity of those last three runs must be 75%*.89 or 66.75%. But we don't yet know how they break down. It may be close to .6675/3 but it can't be exact because of limited permutations which improve either hand. That's why I said using the binomial is an estimate. We either have to do a simulation, or count up the 990*903 second runs to figure the equity on that one, and the 990*903*881 third runs, and the 990*903*881*741 fourth runs. It would be very cumbersome to try to calculate these, and I think a simulation is the only realistic way. Once we know the equity breakdown, then the distribution of winning 0,1,2,3, or 4 times can be calculated accurately.

All that said, I still think the binomial method used in the OP is a good estimate for the long run average. We'd need to figure the standard deviation to know our confidence interval on this one trial.

Sigh, ok after another cup of coffee I changed my mind. I believe the OP method is exactly correct. The average equity must be equal between the runs over time, just not on any specific run. So the binomial distribution works.

But what if you have AsAc against KsKc on a Kh7d2c board?
Your equity is 8.586% and you only have 2 outs so if we run it n times the probability of getting more than 2 wins has to be 0, but if you use the binomial distribution with p=0.08586 you will get some non-zero probability for winning more than 2 runs.

That's an excellent example, and it takes me back to what I said in post #4.

if i wanted to actually run a simulation is there a web page or program which could do this?

KK wins for the following turn/river cards:

KK - 1 way
Kx - where x is not an K or 9 - 84 ways
QJ - 16 ways
TT - 3 ways

Out of the 990 possible. KK and TT can happen only once. Kx can happen twice, if there is no KK. QJ can happen four times. The possible combinations are:

0 - None
1 - One KK, one Kx, one QJ, one TT
2 - KK/QJ, KK/TT, Kx/Kx, Kx/QJ, Kx/TT, QJ/QJ, QJ/TT
3 - KK/QJ/QJ, KK/QJ/TT, Kx/Kx/QJ, Kx/Kx/TT, Kx/QJ/QJ, Kx/QJ/TT, QJ/QJ/QJ, QJ/QJ/TT
4 - KK/QJ/QJ/QJ, KK/QJ/QJ/TT, Kx/Kx/QJ/QJ, Kx/Kx/QJ/TT, Kx/QJ/QJ/QJ, Kx/QJ/QJ/TT, QJ/QJ/QJ/QJ, QJ/QJ/QJ/TT

It's a bit tedious to compute all probabilities but not hard with a computer.